Avoid flood in the city

Array Hash Table Binary Search Greedy Heap (Priority Queue)

39.0% acceptance

Feb 25, 2026

2146

615

Your country has 10^9 lakes. Initially, all the lakes are empty, but when it rains over the nth lake, the nth lake becomes full of water. If it rains over a lake that is full of water, there will be a flood. Your goal is to avoid floods in any lake. Given an integer array rains where: rains[i] > 0 means there will be rains over the rains[i] lake. rains[i] == 0 means there are no rains this day and you must choose one lake this day and dry it. Return an array ans where: ans[i] == -1 if rains[i] > 0. ans[i] is the lake you choose to dry in the ith day if rains[i] == 0. If there are multiple valid answers return any of them. If it is impossible to avoid flood return an empty array.

Solution

Rust

Time O(n log n)

Space O(n)

LeetCode

solution.rs

use std::collections::{BTreeSet, HashMap};

impl Solution {
  pub fn avoid_flood(rains: Vec<i32>) -> Vec<i32> {
    let n = rains.len();
    let mut ans = vec![1i32; n];
    let mut full: HashMap<i32, usize> = HashMap::new(); // lake -> day it was filled
    let mut dry_days: BTreeSet<usize> = BTreeSet::new();

    for (i, &lake) in rains.iter().enumerate() {
      if lake == 0 {
        dry_days.insert(i);
      } else {
        ans[i] = -1;
        if let Some(&fill_day) = full.get(&lake) {
          // Need a dry day after fill_day and before i
          if let Some(&d) = dry_days.range(fill_day + 1..).next() {
            if d < i {
              ans[d] = lake;
              dry_days.remove(&d);
            } else {
              return vec![];
            }
          } else {
            return vec![];
          }
        }
        full.insert(lake, i);
      }
    }
    ans
  }
}